Page 403 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

ΜΛ παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ whole of CL is equal to the (sum of the squares) on AG
τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον and GB, of which CE is equal to the (square) on AB,
ἐστὶ καί ἐστιν ἴσον τῷ ΖΛ, καὶ τὸ ΖΛ ἄρα μέσον ἐστίν. καὶ the remainder FL is thus equal to twice the (rectangle
παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν ΖΜ· ῥητὴ contained) by AG and GB [Prop. 2.7]. Therefore, let
ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ FM have been cut in half at point N. And let NO have
μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ῥητόν ἐστιν, τὸ been drawn through N, parallel to either of CD or ML.
δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον, ἀσύμμετρα [ἄρα] ἐστὶ τὰ ἀπὸ Thus, FO and NL are each equal to the (rectangle con-
τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. ἴσον δέ [ἐστι] τὸ ΓΛ tained) by AG and GB. And since twice the (rectangle
τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ contained) by AG and GB is medial, and is equal to FL,
ΖΛ· ἀσύμμετρον ἄρα [ἐστὶ] τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς FL is thus also medial. And it is applied to the ratio-
τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΜΖ· ἀσύμμετρος ἄρα nal (straight-line) FE, producing FM as breadth. Thus,
ἐστὶν ἡ ΓΜ τῇ ΜΖ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα FM is rational, and incommensurable in length with CD
ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα [Prop. 10.22]. And since the sum of the (squares) on AG
ἐστὶν ἡ ΓΖ. λέγω [δή], ὅτι καὶ τετάρτη. and GB is rational, and twice the (rectangle contained)
᾿Επεὶ γὰρ αἱ ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι, ἀσύμμετ- by AG and GB medial, the (sum of the squares) on AG
ρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ. καί ἐστι τῷ and GB is [thus] incommensurable with twice the (rect-
μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ angle contained) by AG and GB. And CL (is) equal to
ΚΛ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς the (sum of the squares) on AG and GB, and FL equal
τὸ ΚΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΚΜ· ἀσύμμετρος ἄρα to twice the (rectangle contained) by AG and GB. CL
ἐστὶν ἡ ΓΚ τῇ ΚΜ μήκει. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ [is] thus incommensurable with FL. And as CL (is) to
μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστιν ἴσον FL, so CM is to MF [Prop. 6.1]. CM is thus incommen-
τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ἀπὸ τῆς ΗΒ τῷ ΚΛ, surable in length with MF [Prop. 10.11]. And both are
τὸ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τῷ ΝΛ, τῶν ἄρα ΓΘ, ΚΛ μέσον rational (straight-lines). Thus, CM and MF are rational
ἀνάλογόν ἐστι τὸ ΝΛ· ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, (straight-lines which are) commensurable in square only.
οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾿ ὡς μὲν τὸ ΓΘ πρὸς τὸ CF is thus an apotome [Prop. 10.73]. [So], I say that (it
ΝΛ, οὕτως ἐστίν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς is) also a fourth (apotome).
τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ· ὡς ἄρα ἡ ΓΚ For since AG and GB are incommensurable in square,
πρὸς τὴν ΜΝ, οὕτως ἐστὶν ἡ ΜΝ πρὸς τὴν ΚΜ· τὸ ἄρα the (square) on AG (is) thus also incommensurable with
ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ the (square) on GB. And CH is equal to the (square) on
τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί AG, and KL equal to the (square) on GB. Thus, CH is
εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετράρτῳ μέρει τοῦ ἀπὸ τῆς ΜΖ incommensurable with KL. And as CH (is) to KL, so
ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ CK is to KM [Prop. 6.1]. CK is thus incommensurable
τὸ ὑπὸ τῶν ΓΚ, ΚΜ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα in length with KM [Prop. 10.11]. And since the (rectan-
ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί gle contained) by AG and GB is the mean proportional
ἐστιν ὅλη ἡ ΓΜ σύμμετρος μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ· to the (squares) on AG and GB [Prop. 10.21 lem.], and
ἡ ἄρα ΓΖ ἀποτομή ἐστι τετάρτη. the (square) on AG is equal to CH, and the (square)
Τὸ ἄρα ἀπὸ ἐλάσσονος καὶ τὰ ἑξῆς. on GB to KL, and the (rectangle contained) by AG and
GB to NL, NL is thus the mean proportional to CH and
KL. Thus, as CH is to NL, so NL (is) to KL. But,
as CH (is) to NL, so CK is to NM, and as NL (is) to
KL, so NM is to KM [Prop. 6.1]. Thus, as CK (is) to
MN, so MN is to KM [Prop. 5.11]. The (rectangle con-
tained) by CK and KM is thus equal to the (square) on
MN—that is to say, to the fourth part of the (square) on
FM [Prop. 6.17]. Therefore, since CM and MF are two
unequal straight-lines, and the (rectangle contained) by
CK and KM, equal to the fourth part of the (square) on
MF, has been applied to CM, falling short by a square
figure, and divides it into incommensurable (parts), the
square on CM is thus greater than (the square on) MF
by the (square) on (some straight-line) incommensurable

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